Integrated differential amplifier 1-1 Difference Voltage

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1 Integrated differential amplifier 1-1 Difference Voltage A differential amplifier is also called a difference amplifier, because it amplifies the difference between two signal voltages, (a difference voltage is the mathematical difference between two other voltages, each of whose values is given with respect to ground), for example, the collector-to-emitter voltage of a BJT is the difference between the collector-to- ground voltage and the emitter-to-ground voltage: VCE = VC – VE---------------------------------------------1-1 Fig1-1 The amplification of difference voltages The two input voltages is v1 & v2, are shown as sine waves. the voltage gain of each amplifier is A, then vo1=Av1 & vo2= Av2 . The input difference voltage, v12 = v1 - v2, is a sine wave . the output difference voltage, Av1 - Av2 = A(v1 - v2). 1-2 The Ideal Differential Amplifier Fig1-2The basic BJT differential amplifier, the two transistors can be regard as CE amplifiers having a common connection at their emitter. The base terminals are the input to the differential and the collectors are the outputs

Transcript of Integrated differential amplifier 1-1 Difference Voltage

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Integrated differential amplifier1-1 Difference VoltageA differential amplifier is also called a difference amplifier, because it amplifies the difference between two signal voltages, (a difference voltage is the mathematical difference between two other voltages, each of whose values is given with respect to ground), for example, the collector-to-emitter voltage of a BJT is the difference between the collector-to-ground voltage and the emitter-to-ground voltage:

VCE = VC – VE---------------------------------------------1-1

Fig1-1 The amplification of difference voltagesThe two input voltages is v1 & v2, are shown as sine waves. the voltage gain of each amplifier is A, then vo1=Av1 & vo2= Av2 . The input difference voltage, v12 = v1 - v2, is a sine wave . the output difference voltage, Av1 - Av2 = A(v1 - v2).1-2 The Ideal Differential Amplifier

Fig1-2The basic BJT differential amplifier, the two transistors can be regard as CE amplifiers having a common connection at their emitter. The base terminals are the input to

the differential and the collectors are the outputs

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Figure 1-4 shows the amplifier with input 2 grounded (vi2 = 0) and a small signal applied to input 1. The ideal current source presents an infinite impedance (open circuit) to an ac signal. We also assume the ideal situation of perfectly matched transistors, so Q1 and Q2have identical values of β, re . Since Q1 is CE amplifier, the voltage at its collector (vo1) is an amplified and inverted version of its input vi1. there is also an ac voltage ve1 developed at the emitter of Q1 this voltage is in phase with vi1

Fig1-3 Schematic symbol for the differential amplifier

Fig1-4 The small signal voltages in a differential amplifier when one input is grounded, Note that ve1 is in phase with vi1 and vo1 is out of phase with vi1

The voltage ve1 is developed across the emitter resistance re looking into the emitter of Q2(in parallel with the infinite resistance of the current source). Therefore, as far as the emitter-follower action of Q1 is concerned, the load resistance seen by Q1 is re . Since the emitter resistance of Q1 is itself re, then the emitter-follower gain is:

ve1 is in phase with, and 1/2 the magnitude of vi1.vbe2 = vb2 – ve1 = 0 – ve1, We see that even though the base of Q2 is grounded, there exists an ac base-to-emitter voltage on Q2 that is out of phase with ve1 and therefore out of phase with vi1.Consequently, there is an ac output voltage vo2 produced at the collector of Q2 and it is out of phase with vo1,Since both transistors are identical, they have equal gain and the output vo2 has the same magnitude as vo1. Since the emitter-follower gain of Q1 is 0.5,

ve is a 0.5(100 mV) = 50-mV-peak sine wave.

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Fig1-5 Each transistor has identical voltage gain( -100) and the outputs at the collectors are -100 times their respective base-to-emitter voltage

vbe1= vb1 – ve1 = (100 mV) - (50 mV) = 50 mV.vo1 = A x vbe1= -100(50 mV) = -5 V, an inverted 5-V-peak sine wave.vbe2= vb2 - ve1 = 0 - (50 mV) = -50 mV .vo2 =A x vbe2 = (-100)( - 50 mV) = + 5 V peak sine wave in phase with vi1 and out of phase with vo1.the input difference voltage is vi1 – vi2 = (100 mV) - 0 = 100 mV peak. the output difference voltage is 10V peak, since vo1 and vo2 are out of phase.the magnitude of the difference voltage gain (vo1 - vo2)/(vi1 - vi2) =10 V/ 100 mV= 100. the voltage gain vo/vi for each side is only 50, the difference voltage gain is the same as the gain vc/vbe of each transistorthe two inputs of a differential amplifier are driven by signals that are equal in magnitude and out of phase: vi2 = -vi1.let us now ground input 1 (vi1 = 0) and assume that there is a signal applied to input 2 equal to and out of phase with the vi1 signal we previously assumed. vo2 is out of phase with vi1and vo1 is in phase with vi2. In figure 1-7, driving the two inputs with equal but out-of-phase signals . By superposition, each output is the sum of the voltages resulting from each input acting alone, so the outputs are exactly twice the level they would be if only one input signal were present. In many applications, the output of a differential amplifier is taken from just one of the transistor collectors, vo1, for example. In this case the input is a difference voltage and the output is a voltage with respect to ground. This called single-ended output and the voltage gain in that mode is

------1-2

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the single-ended output gain is one-half the difference voltage gain (vo1 - vo2)/ (vi1 – vi2) which called as the double-ended voltage gain.

Fig1-6 The differential with vi1 grounded and a signal input vi2, campier with fig1-4 , Note that vi2 here is the opposite phase from vi1 in fig1-4 and that vo1 and vo2 are the same as in

fig1-4

Fig1-7By the superposition principle, the output vo1 when both input are applied is the sum of the vo1 outputs due to each signal acting alone, likewise for vo2

Example 1-1. The magnitude of the voltage gain (vc/vbe) for each transistor in Figure 1-2 is 100. If vi1 and vi2 are out-of-phase, 100-mV-peak signals applied simultaneously to the inputs, findI. the peak values of vo1 and v02,

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2. the magnitude of the double-ended voltage gain (vo1 - vo2)/(vi1 – vi2), and3. the magnitude of the single-ended output gain vo1/(vi1 – vi2).Solution.1. the peak value of each output = A x vbe1= 100 x 50 mV = 5V when one input is driven and the other is grounded. Since the outputs are doubled when the inputs are equal and out of phase,

each output = 2 x 5 V = 10V peak.2. Since vi1 = -vi2, the input difference voltage is vi1 - vi2 = 2vi1 = 200 mV peak.vo1 = -vo2, so the output difference voltage is vo1 - vo2 = 2vo1 = 20 V peak . Therefore, the magnitude of (vo1 - vo2)/(vi1-vi2) = (20 V)/(200 mV)= 100.3. The magnitude of the single-ended output gain is

vo1 is out of phase with (vi1 - vi2), the correct specification for the single-ended output gain is -50. If the single-ended output is taken from the other side (vo2), which is out of phase with vo1, then the gain vo2/(vi1- vi2) = +50.

Fig1-8 the output of the differential amplifier are 0 when the two inputs are equal and in phase

Since the output difference voltage vo1 - vo2 is out of phase with the input difference voltage vi1 - vi2, the correct specification for the double-ended voltage gain is -100.if the two inputs are driven by equal in-phase signals, the output at each collector will be exactly 0, and the output difference voltage will be 0, the input difference voltage is also 0. We are again assuming that the current source has infinite resistance. Neglecting the output resistance ro at the collector of Q1, we can use the familiar approximation for the voltage gain of the transistor:

------------------------1-3

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Fig1-9 when the input to Q2 is grounded, there is resistance re in series with the emitter of Q1

where re is the emitter resistance of Q1. It is clear from Figure 1-9 that the voltage gain vo1/vi1 is

------------------------1-42re is in the denominator because we assume re1=re2, i.e. double-ended voltage gain

-----------------------------1-5single-ended output voltage gain

-----------------------------1-6that vo1 and vo2 will always have the same amplitude and be out of phase with each other. Thus,

-----------------------------1-7The small-signal differential input resistance is defined to be the input difference voltage divided by the total input current. Since the total small-signal resistance in the path from one input through both emitters to the other input is 2re, the differential input resistance is

------------------------1-8Since the transistors are identical, the source current I divides equally between them, and the emitter current in each is therefore

------------------1-9The dc output voltage at the collector of each transistor is

-------------1-10the familiar approximation re ≈ 0.026/IE we obtain

---------1-11

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Fig1-10 DC voltage and currents in an ideal differential Amp Fig1-11 Example1-2

Example 1-2. For the ideal differential amplifier shown in Figure 1-11, find 1. the dc output voltages vo1 and vo2,2. the single-ended output gain vo1/(vi1 – vi2), and3. the double-ended gain (vo1 - vo2)/(vi1 – vi2).Solution.1. The emitter current in each transistor is IE = I/2 = (2 mA)/2 = 1 mA ≈ Ic. Therefore,

vo1 = vo2 = Vcc - IcRc = 15 - (1 mA)(6 kΩ) = 9 V.2. The emitter resistance of each transistor is

Therefore, from equation 1-6

3.from equation1-5

1-2 Common-Mode ParametersOne attractive feature of a differential amplifier is its ability to reject signals that are common to both inputs. Since the outputs are amplified versions of the difference between the inputs, any voltage component that appears identically in both signal inputs will be “differenced out" that is, will have zero level in the outputs. Any dc or ac voltage that appears simultaneously in both signal inputs is called a common-mode signal vcm. The ability of an amplifier to suppress, or zero-out, common-mode signals is called common-mode rejection. The differential common-mode gain Acm, is defined to be the ratio of the output difference voltage caused by the common-mode signal to the common-mode signal itself:

--------------1-12

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Fig1-12 if the differential amplifier were ideal, both outputs would be 0 when the inputs have the same(common-mode) signal, in reality, there is a small common-mode output, as shown

Obviously the ideal amplifier has common-mode gain equal to 0.Common-mode rejection ratio (CMRR), defined to be the ratio of the magnitude of its -differential (difference-mode) gain Ad to the magnitude of its common-mode gain

--------------------------1-13The value of the CMRR is often given in decibels:

--------------1-14

Example 1-3. When the inputs to a certain differential amplifier are vi1 = 0.1sin wt and vi2= -0.1 sin wt, it is found that the outputs are vo1 = -5 sin wt and vo2 = 5 sin wt. When both inputs are 2 sin wt, the outputs are vo1 = -0.05 sin wt and vo2 = 0.05 sin wt. find the CMRR in dB.Solution. We will use the peak values of the various signals for our gain computations, but note carefully how the minus signs are used to preserve phase relations , the difference-mode gain is

The common-mode gain is

The common-mode rejection ratio is

Expressing this result in dB, we have CMRR

1-3 Practical Differential AmplifierOur gain derivations for the ideal differential amplifier were based on the assumption that both transistors had identical values of re Clearly, the voltage gains of both sides will not be identical if the values of re are not, in which case the outputs will not truly represent amplified versions of the input difference voltage, and the CMRR will suffer. To reduce the

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effect of variations in re, equal-valued resistors RE can be inserted in series with the emitters, as shown in Figure 1-13.

Fig1-13 Inserting a resistance RE in series with each emitter reduces the amplifiers dependence on matched re values

Equations 1-5 and 1-6, modified for the inclusion of RE, become

----------------------------1-15

------------------1-16Equation 1-8 becomes

-----------------------------1-17Figure 1-14 shown Another reality in practical differential amplifiers is that the current source biasing the amplifier does not have infinite resistance, the current-source resistance in each half-circuit must be 2R and the value of the current must be I/2 to maintain equivalence.The voltage gain of Q1 is

-------------------------------1-18And that of Q2 is

--------------------------------1-19Therefore

This shows that the differential common-mode gain is unaffected by source resistance R .

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Fig1-14 Analyzing the effect of source resistance R on common-mode behavior

1-4 Bias Methods in Integrated Circuitsintegrated-circuit amplifiers use transistor constant-current sources, an example of which is illustrated in Figure 1-15. Transistor Q3 has a large output resistance at its collector therefore assuming that (βRE3 )>>( R1 R2), the base voltage of Q3 is

--------------1-20Assuming a silicon transistor, the emitter voltage of Q3 is

---------------1-21

--------------------1-22Example1-4 Transistor Q3 in fig1-16 has β=100.Assuming that Q1 and Q2 are matched, find approximate values for1-the emitter current in Q1 and Q2, and2-the dc output voltage vo1 and vo2Solution. 1-

2-

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Fig1-15 a transistor current source fig1-16 example(1- 5) used to bias a differential Amp

Example1-5 Assuming that each of Q1 and Q2 in Figure 1-16 has β =100 and that the output resistance at the collector of Q3 is 500 kΩ, find1. the differential input resistance,2. the single-ended common-mode gain, and3. the single-ended common-mode rejection ratio.Solution: The small-signal emitter resistance of Q1 and Q2 is1- the small-signal emitter resistance of Q1 and Q2 is

Equations 1-18 &1-19 are for the case RE =0. When resistance RE is included each emittercircuit, the single-ended common-mode gain is

3. from equation 1-16 the single-ended output gain is

Therefore, the single-ended CMRR is

Or 70.3 dB1-6 Introduction to Operational AmplifiersAn operational amplifier is basically a differential amplifier modified by the addition of circuitry that improves its performance and gives it certain special features. The mostimportant characteristics of an operational amplifier are listed below:

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1. It is a dc (direct-coupled, direct-current) amplifier.2. It should have a very large voltage gain-ideally, infinite.3. It should have a very large input impedance-ideally, infinite.4. It should have a very small output impedance-ideally zero.5. The output should be exactly zero V when the inputs are zero V.6. The output must be capable of both positive and negative voltage swings.7. It should have a very large CMRR.8. It is operated with a single-ended output and differential input (although one input is often grounded,).9. It should meet whatever special requirements are demanded by a particular application these include parameters such as noise level, frequency response, and slew rate.The name operational amplifier is derived from amplifier applications that the performance of precise mathematical operations on input signals, including voltage summation, subtraction, and integration. The input stage of every operational amplifier is a differential amplifier. To achieve a large input impedance, Components in the input stage should be very closely matched to achieve the best possible balance in the differential operation. This is important to ensure that the output of the operational amplifier is a precise representation of the input difference voltage, that the output is exactly zero when the inputs are zero, and that the CMRR is large. voltagegain is achieved through the use of multistage amplifier. To permit the output voltage to swing through both positive and negative values, both a positive and a negative supply voltage are required. These are usually equal-valued, opposite-polarity supplies, a typical example being ± 15 V.

Circuit Analysis of an Operational AmplifierExample1-6 Figure 1-17 shows a simple operational amplifier that we can use as an example to analyzed important function components discussed

VE7= V B7 - 0.7 = - 10.9 V,

IC1 = IC2 = IE1= IE2 = (0.4 mA)/2 = 0.2 mA,VC1=VC2= Vcc -IcRc = 15 - (0.2 mA)(25 kΩ) = 10 V.

VE1=VE2= 0 - 0.7 = -0.7 V,the small drop across each 50-Ω resistor [(50 ) x (0.2 mA) = 0.01 V] sets the collector of Q7at about the same voltage (-0.71 V).Since VB8 = VB7 = -10.2 V, i.e. VE8= VB8 - 0.7 = -10.9 V. Then

The 1.8 mA divides equally between Q3 and Q4, so VC4 = Vcc - IcRc = 15 - (0.9 mA)(3.3kΩ) = 12 V. Since the bases of Q3 and Q4 are direct-coupled to the collectors of Q1 and Q2,

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Figure 1-17 A simple operational amplifier incorporating deferential, gain. And level-shifting stages. All voltages shown are dc levels with respect to ground

VB3 = VB4 = 10 V.VE3 = VE4 = 10 - 0.7 = 9.3 V.

VB5 = 12 V. Therefore, VE5 = VB5 + 0.7 = 12.7 V.

Since IC5 = IE5, VC5 = (IC5)(10.47 kΩ) - VEE = (1.5 mA)(10.47 kΩ) - 15 = +0.7 V.Since the base of the output transistor, Q6, is at 0.7 V, its emitter is at 0 V, and the amplifier output is 0 V. The bias current in Q6 is (0 - VEE)/5 kΩ = (15 V)/ (5 kΩ) = 3 mA.

Example 1-7. Assume that the transistors in Figure 1-17 are matched and that all have β=100. Neglecting the collector output resistance of each transistor, find1. the voltage gain Vo/(Vi1 – Vi2),2. the differential input resistance of the amplifier,3. the output resistance of the amplifier.Solution.

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1. The load driven by the input differential stage is the differential input resistance rid34 ofthe second stage. Since IE3 = IE4 = 0.9 mA,

Therefore, rid34 = β (re3 + re4) = 100(28.9 + 28.9) = 5.78 kΩ.

The ac equivalent circuit of the first stage is shown in Figure 1-18. The double ended voltage gain is given by

-------1-23

The ac load resistance driven by the second stage is the input resistance looking into the base of Q5:

ri5 = β (re5 + RE5). Since IE5 = 1.5 mA, re5 = 0.026/(1.5 mA) = 17.3 Ω.Thus, ri5 = 100[(17.3 Ω) + (1.53 kΩ)] = 154.73 kΩ.

The second stage is operated single-ended and its gain is

Figure 1-8 (Example 1-7) The ac equivalent circuit of the input stage,showing the differential input resistance of the second stage connected between

the collectors

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The resistance in the collector circuit of the level-shifting stage (Q5) is (10.47 kΩ) ri6, where ri6 is the input resistance looking into the base of Q6,

Since IE6 = 3 mA, re6 = 0.026/(3 mA) = 8.70and ri6 = β (re6 + RE6) = 100[(8.70)+ (5 kΩ)] = 500 kΩ.

Thus, the gain of Q5 is

The overall gain of the amplifier is the product of the gain calculated for the stages

(This would not be considered a very large voltage gain for modem operational amplifiers.)2. The differential resistance looking into the first stage is

rid12 = β (re1 + re2 + 2RE) = 100(130 + 130 + 100) = 36 kΩ.3. Recall that the output resistance of an emitter-follower stage is